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nC0 = 1which says "no one can be selected out of n objects in just one way".But I can't imagine this thing of "no one can be selected in one way".So,I want you to give a proper real life example to help me think in a better way.

user87284
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  • $\binom{n}{0}$ is equal to $\frac{n!}{0!\cdot n!}$. You can select $n$ elements in one order only like $\underbrace{aa\ldots aa}_{\textrm{n-times}}$. There is no way where you can select elements where the element $b$ is involved. – callculus42 May 21 '20 at 16:42
  • Sequences like coinflips are fine for building intuition... Consider flipping a coin three times. The following sequences are possible: $\underbrace{TTT}{\binom{3}{0}=1},\underbrace{TTH,THT,HTT}{\binom{3}{1}=3},\underbrace{THH,HTH,HHT}{\binom{3}{2}=3},\underbrace{HHH}{\binom{3}{3}=1}$ where we organized things based on the size of the subset of the positions which are heads. The sequence corresponding to where the subset of positions occupied by heads being empty corresponded to the sequence $TTT$ of which there is $\binom{3}{0}=1$ such option. – JMoravitz May 21 '20 at 16:54
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    In terms of people being selected... consider a class of $n$ people. They take a test. Anyone who passes the test gets to enjoy their weekend. Anyone who failed the test has to stay for supplementary lessons. Consider the different ways in which the students might either get to enjoy their weekend or stay behind for extra lessons. Included among those ways is where all of the students had to stay behind (i.e. the collection of students who passed the test was empty) or alternatively where all of the students got to enjoy their weekend (the collection of students who failed was empty) – JMoravitz May 21 '20 at 16:58

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Consider a set of four objects $\{A, B, C, D\}$. To pick $n$ objects from that set is to divide it into two sets: the set of picked objects, and the set of unpicked objects. There are $4$ ways to point $1$ object:

$$\begin{array}{c|c}\text{Picked}&\text{Unpicked}\\\hline \{A\}&\{B, C, D\}\\\{B\}&\{A, C, D\}\\\{C\}&\{A, B, D\}\\\{D\}&\{A, B, C\}\end{array}$$ So ${4\choose 1} = 4$.

There are $6$ ways to pick $2$ objects:

$$\begin{array}{c|c}\text{Picked}&\text{Unpicked}\\\hline \{A, B\}&\{C, D\}\\\{A, C\}&\{B, D\}\\\{A, D\}&\{B, C\}\\\{B, C\}&\{A, D\}\\\{B, D\}&\{A, C\}\\\{C, D\}&\{A, B\}\end{array}$$ So ${4\choose 2} = 6$.

I could list the sets for picking $3$ objects, in fact I already have: just swap the columns on the "pick $1$ object" table. This is why ${4 \choose 3} = {4 \choose 1} = 4$.

So how can you pick $0$ objects? That means that the set of picked objects is empty: $$\begin{array}{c|c}\text{Picked}&\text{Unpicked}\\\hline \{\}&\{A, B, C, D\}\end{array}$$

There is only one empty set, so there are no other options for picking $0$ objects. And just as with pick $1$ vs pick $3$, pick $0$ turns into pick $4$ when you swap columns. There is also only one way to pick $4$ items: you have to pick them all. So ${4 \choose 0} = {4 \choose 4} = 1$.

Paul Sinclair
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